\newproblem{lay:5_5_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.5.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be an $n\times n$ real matrix with the property that $A^T=A$. Show that if $A\mathbf{x}=\lambda\mathbf{x}$ for some nonzero vector
	$\mathbf{x}\in\mathbb{C}^n$, then, in fact, $\lambda$ is real and the real part of $\mathbf{x}$ is an eigenvector of $A$. [\textit{Hint}: Compute
	$(\mathbf{x}^*)^TA\mathbf{x}$ and use Exercise 5.5.23. Also, examine the real part of $A\mathbf{x}$.]
}{
  % Solution
	Let us calculate $q=(\mathbf{x}^*)^TA\mathbf{x}$
	\begin{center}
		$\begin{array}{rcl}
			q&=&(\mathbf{x}^*)^TA\mathbf{x}=(\mathbf{x}^*)^T(A\mathbf{x})=(\mathbf{x}^*)^T(\lambda\mathbf{x})=\lambda\|\mathbf{x}\|^2
		\end{array}$
	\end{center}
	Since $\|\mathbf{x}\|^2$ is a real number and $q$ is a real number, then $\lambda$ is a real number.
	
	Let us calculate the real part on both sides of the equation $A\mathbf{x}=\lambda\mathbf{x}$
	\begin{center}
	  $\begin{array}{rcll}
		   \mathrm{Real}\{A\mathbf{x}\}&=&\mathrm{Real}\{\lambda\mathbf{x}\}&\text{[A and $\lambda$ are real]} \\
		   A\mathrm{Real}\{\mathbf{x}\}&=&\lambda\mathrm{Real}\{\mathbf{x}\}& \\
		\end{array}$
	\end{center}
	So $\mathrm{Real}\{\mathbf{x}\}$ is an eigenvector of $A$.
}
\useproblem{lay:5_5_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
